MATH SOLVE

5 months ago

Q:
# Chuck planted an unusual orange tree one afternoon. it came with 4 oranges on it and the number of oranges on the tree doubles each morning. chuck, upon returning home for lunch each day, picks and eats 3 oranges for his lunch. fred has an unusual grapefruit tree. each morning four more grapefruits grow on the tree. his family picks and eats 6 grapefruits at lunchtime. how many grapefruits were on fred's tree in the morning on the day chuck planted his tree, if on the tenth day (in the morning ) the number or oranges and grapefruits, on both trees, were exactly the same

Accepted Solution

A:

Chuck's orange tree describes a geometric progression with the first term of 4 and a common ratio of 2.

On the 10th day, the number of oranges on produced by Chuck's orange tree is given by:

[tex]U_n=ar^n \\ \\ U_{10}=4(2)^{10} \\ \\ =4(1024)=4,096\ oranges[/tex]

Given that Chuck picks 3 oranges everyday for lunch, then in the morning of the 10th day, he must have picked 3 oranges for 9 days making the total number of oranges picked by Chuck for lunch to be 3(9) = 27.

Thus, the number of oranges remaining on the orange tree on the 10th day morning is given by 4,096 - 27 = 4,069.

Let the number of grapefruits on Fred's grape tree on the morning Fred planted his orange tree be x. Fred's grape tree's production describes an arithmetic progression with the first term of x and a common difference of 4.

On the 10th day, the number of grapefruits produced by Fred's grape tree is given by:

[tex]U_n=a+(n-1)d \\ \\ U_{10}=x+(10-1)4 \\ \\ =x+9(4)=x+36[/tex]

Given that Fred's family picks 6 grapefruits everyday for lunch, then in the morning of the 10th day, he must have picked 6 grapefruits for 9 days making the total number of grapefruits picked by Fred's family for lunch to be 6(9) = 54.

Thus, the number of grapefruits remaining on the grape tree on the 10th day morning is given by x + 36 - 54 = x - 18.

Given that the number or oranges and grapefruits, on both trees, were exactly the same on the 10th day's morning, then

x - 18 = 4,069

x = 4,069 + 18 = 4,087

Therefore, the number of grapefruits on Fred's grape tree on the morning Fred planted his orange tree was 4,087 grapefruits.

On the 10th day, the number of oranges on produced by Chuck's orange tree is given by:

[tex]U_n=ar^n \\ \\ U_{10}=4(2)^{10} \\ \\ =4(1024)=4,096\ oranges[/tex]

Given that Chuck picks 3 oranges everyday for lunch, then in the morning of the 10th day, he must have picked 3 oranges for 9 days making the total number of oranges picked by Chuck for lunch to be 3(9) = 27.

Thus, the number of oranges remaining on the orange tree on the 10th day morning is given by 4,096 - 27 = 4,069.

Let the number of grapefruits on Fred's grape tree on the morning Fred planted his orange tree be x. Fred's grape tree's production describes an arithmetic progression with the first term of x and a common difference of 4.

On the 10th day, the number of grapefruits produced by Fred's grape tree is given by:

[tex]U_n=a+(n-1)d \\ \\ U_{10}=x+(10-1)4 \\ \\ =x+9(4)=x+36[/tex]

Given that Fred's family picks 6 grapefruits everyday for lunch, then in the morning of the 10th day, he must have picked 6 grapefruits for 9 days making the total number of grapefruits picked by Fred's family for lunch to be 6(9) = 54.

Thus, the number of grapefruits remaining on the grape tree on the 10th day morning is given by x + 36 - 54 = x - 18.

Given that the number or oranges and grapefruits, on both trees, were exactly the same on the 10th day's morning, then

x - 18 = 4,069

x = 4,069 + 18 = 4,087

Therefore, the number of grapefruits on Fred's grape tree on the morning Fred planted his orange tree was 4,087 grapefruits.